v^2=-42+13v

Solution for v^2=-42+13v equation:

v^2=-42+13v

We move all terms to the left:

v^2-(-42+13v)=0

We add all the numbers together, and all the variables

v^2-(13v-42)=0

We get rid of parentheses

v^2-13v+42=0

a = 1; b = -13; c = +42;
Δ = b2-4ac
Δ = -132-4·1·42
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-1}{2*1}=\frac{12}{2}
=6
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+1}{2*1}=\frac{14}{2}
=7
$